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Point) Let[3] y = ~10For what value of h is y in the span of the vectors V1 ad V2 h =...

Question

Point) Let[3] y = ~10For what value of h is y in the span of the vectors V1 ad V2 h =

point) Let [3] y = ~10 For what value of h is y in the span of the vectors V1 ad V2 h =



Answers

Given the vectors $\mathbf{P}=3 \mathbf{i}-\mathbf{j}+\mathbf{k}, \mathbf{Q}=4 \mathbf{i}+Q, \mathbf{j}-2 \mathbf{k},$ and $\mathbf{S}=2 \mathbf{i}-2 \mathbf{j}+2 \mathbf{k},$ determine the value of $Q_{y}$ for which the three vectors are coplanar.

For this exercise. We are provided with three different vectors, and the vector V three we see here also depends on the value of age. So questions we might ask about this particular set of vectors are for what values of age would this set be linearly dependent? Well, to answer that kind of question, we can start by making a Matrix A, which is formed using the columns V. One, V two and V three are provided vectors. Then our metric say would be 15 negative three for the first column. Negative to negative 96 and call him, too, and three h negative nine for the third column. Let's re reduce this toe echelon form. That way we can analyze the pivots and now determined linear dependence. So let's begin then with Row one or entry here our first pivot position and will take out the five and the negative three below. So let me copy one negative two and three for a first step. Multiply row one by negative five. Added to row two, we obtained 01 and H minus 15 for the next step. If we triple row one multiplied by three and after Row three. Then we'll produce a zero zero and zero as well. So that's kind of convenient. With one row operation, we are in echelon form now. The pivots will be here and here and nowhere else. And the value of H does not change where the pivots will be. Let's consider that matrix equation eight times X equals a zero vector. This equation has nontrivial solutions. The reason we know this is that there is a free variable that corresponds to the third column in this particular matrix. So since we have non trivial solutions, we know nontrivial solutions to a homogeneous matrix equation implies immediately that the set of vectors V one, V two and V three is linearly dependent, and the linear dependence we found again does not depend on H whatsoever. So V one v two V three is linearly dependent for all values h in the set of Real's. So this is how we can determine whether or not we have linear, independent or dependent vectors in this situation,

Okay, so for a problem 18 we wouldn't want to find the value of age. So that, uh, Victor why he's saying that he's on the plane generated to buy of You want to be too. So, by the giving information we know first that next next to Sorry, I'll just first right down. Do you want me to? So you wanna be too? Will be 10 connected to and neck of 318 And Victor, why will be age next five and collective three. So that is actually to consider this Al Commended Matrix age Negative five. Elective three. We want this mission this matrix to be consistent. That is the say, a matrix v spend by spend by V one and V two. So we're considering the system. He actually was. Why? Okay, So we need to do what we need to do here is to performed the cash indignation. So it's not a complicated matri. So let's do this by hand. So way first used just 1/3 row plus twice up the first rule. So row three, us twice off row one. So we can cancel the first entry on the third row. So He's a general and connective six here and plus eight. So we'll be inactive. Started positive, too. And for the constant term, we have connective three plus two age. Okay. And I'll keep the kitty out of bettors. One negative three and a JJ. Okay, so the next we want to cancel the third the second entry off on 1/3 row. So to do that, we we use Ah, the role minus twice off the second roll. So, bro. Three minus twice off road, too. So it will be one first. Right down. Ian changed those, and we have 01 and negative five. So for the last roll, there are zero to minus two times one. So that zero and negative three plus two age miners two times negative. Five. So? So it's ah, um, right down. So elective three lasts to age minus two times. Row two will be negative. Five. So here it will be naked. Three plus two age plus 10. And this will be seven thus to age. Okay, so too to make they're all commended me may trace. Um, consistent. We need to make the last entry on the right hand side to be zero. So we require, in this case, age to be inactive. Seven have so that day sentry will be zero than the system can be solved. Otherwise the system cannot be solved.

Okay. The ocean is and lead and equals two C span. We went way do. Okay. And bases are Leven, and we do. Okay, so we have to show that eggs is in edge and they help to find the basis. Coordinate Tractor off. Coordinate Victor off X. Okay. And given vectors are even is 11 minus five. 10 7 We do is what is given 14 minus eight, 13. 10. And we treat and access. The solution is 19 minus 13, 18 and 15. Okay, so we have to Jack. So then now we will put in the Matrix, actually. Way need to be able to write X equals two x equals toe everyone plus b. We do. In other words, we want to find if the system of linear equation has solution. Okay, so we will port, then put metrics a questo, even we to an X in medlab. Okay, in my lab. And we will use command our CF. Actually, we can solve these weather. Landy matters what? It will be very time consuming so you can use the mat lab as well as the handwriting matter. So the metrics it will be It will be 11 one second it is. I have to show that Yes. 11 minus 5. 10 7. It is even. And 14 minus eight. 13. 10 It is. We do 19 minus 13, 18 and 15. It is the X. When we used the medlab and put the command R E f A, he will get when Zito minus fired by three 018 by 3000000 So, system has a solution first. We know from in this answer they're the system as a solution. So it is proved that X is in edge. Okay? And from the metrics, we can also read the coordinate off. XB is minus five by three and eight by three. And these are final Ansett Final answer. Thank you.

So we're gonna find the dot product of the given vectors. Now, to real quickly, um look at our equations over here, we will be multiplying the X components of each of our vectors, and then we're going to add it to the multiple of the y components of each of the vectors. So here we have two times 3 And then we have negative three times. To that would give us a 6 -6, which is zero. So go ahead and sketch these out, just kind of like a coordinate system, and you're going to see that these are 90° apart, or what you can say is perpendicular.


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